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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    (Original post by Kennethm)
    I got the answer D, if thats correct Ill explain to you
    That is indeed the correct answer. Please, do explain!
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    (Original post by Kennethm)
    I got the answer D, if thats correct Ill explain to you
    can you explain
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    (Original post by bwr19)
    That is indeed the correct answer. Please, do explain!
    literally if r is decreased so that the electric potential energy of Q is doubled then that means r has been halved.

    here is the thought process:
    W = QV and V has been doubled

    V = KQ/r

    if V is doubled then r must have halved.

    referring back to the force mentioned at the start of the question F = KPQ/r^2

    (forget about the constant here) then we have F is proportional to 1/r^2

    so if r is halved then the force will increase by 4 times as (0.5)^2 is 0.25.

    hope this helps.
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    (Original post by Kennethm)
    literally if r is decreased so that the electric potential energy of Q is doubled then that means r has been halved.

    here is the thought process:
    W = QV and V has been doubled

    V = (-)GM/r (dont worry about the minus)

    if V is doubled then r must have halved.

    referring back to the force mentioned at the start of the question F = KPQ/r^2

    (forget about the constant here) then we have F is proportional to 1/r^2

    so if r is halved then the force will increase by 4 times as (0.5)^2 is 0.25.

    hope this helps.
    Hate to be picky - but didn't it mention electrical potential energy, not gravitational?
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    (Original post by CD223)
    Hate to be picky - but didn't it mention electrical potential energy, not gravitational?
    ohhhh thats so embarresing! Same thing nonetheless but ill change it.
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    (Original post by Kennethm)
    ohhhh thats so embarresing! Same thing nonetheless but ill change it.
    not embarassing you were pretty much there.

    for potential to increase by 2, radius must have been halved, as v=1/r

    Force is proportional to 1/r^2.

    F= 1/(1/2)^2. Therefore, 4F.

    is this right?
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    (Original post by wilson196)
    not embarassing you were pretty much there.

    for potential to increase by 2, radius must have been halved, as v=1/r

    Force is proportional to 1/r^2.

    F= 1/(1/2)^2. Therefore, 4F.

    is this right?
    spot on my friend.
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    Hey guys,
    So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
    A The magnitude of the change in momentum of the molecule is zero.
    B The magnitude of the change in momentum of the molecule is 2mu.
    C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
    D The change in kinetic energy of the molecule is zero.

    Whilst the mark scheme gives the answer to be A, the examiner's report claims:
    'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu − (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'

    So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?
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    (Original post by zaxxo1)
    Hey guys,
    So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
    A The magnitude of the change in momentum of the molecule is zero.
    B The magnitude of the change in momentum of the molecule is 2mu.
    C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
    D The change in kinetic energy of the molecule is zero.

    Whilst the mark scheme gives the answer to be A, the examiner's report claims:
    'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu − (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'


    So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?
    I know this question - the answer is out of A and B since they contradict each other. A lot of students would have thought that the momentum to the right would have cancelled out completely with the momentum to the left without fully considering the basic mechanics of the CHANGE in momentum.

    read the question again - it asks for the incorrect answer. B is correct so you must put A down as the answer.
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    (Original post by zaxxo1)
    Hey guys,
    So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
    A The magnitude of the change in momentum of the molecule is zero.
    B The magnitude of the change in momentum of the molecule is 2mu.
    C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
    D The change in kinetic energy of the molecule is zero.

    Whilst the mark scheme gives the answer to be A, the examiner's report claims:
    'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu − (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'

    So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?
    it asks for the incorrect answer, b, c and d are all correct. A is incorrect due to the 2mu calculation disproving this.
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    Not prepared for this at all :/
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    (Original post by gabriellakhan)
    Not prepared for this at all :/
    I was a few weeks ago. Oddly, the closer I get to the exam, and more work I do, the less prepared I feel
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    (Original post by gcsestuff)
    Attachment 424271

    For question 8 I keep getting d, it's c


    Posted from TSR Mobile
    Use E=(1/2)CV^2 for capacitors
    also P = ΔW/Δt

    then you can derive P = ((1/2)CV2)/Δt
    so Name:  CodeCogsEqn.gif
Views: 120
Size:  1.1 KB=3.5x106W=4.5MW
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    (Original post by AR_95)
    I was a few weeks ago. Oddly, the closer I get to the exam, and more work I do, the less prepared I feel
    I 100% feel the same! I've learnt more in these couple of weeks but now I feel like I don't know enough


    Posted from TSR Mobile
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    Name:  ImageUploadedByStudent Room1433942090.728568.jpg
Views: 127
Size:  103.1 KB

    Is this the reasons why things move quicker with a smaller radius? Also why does a satellite need to move quicker to stop it falling into the earth?


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    How is the answer to this C and not B? Surely

    T = 2\pi\sqrt{\frac{l}{2g}}

    so

    \sqrt{2}T = 2\pi\sqrt{\frac{l}{g}}


    ????Name:  Capture.PNG
Views: 129
Size:  37.6 KB
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    (Original post by gcsestuff)
    Name:  ImageUploadedByStudent Room1433942090.728568.jpg
Views: 127
Size:  103.1 KB

    Is this the reasons why things move quicker with a smaller radius? Also why does a satellite need to move quicker to stop it falling into the earth?


    Posted from TSR Mobile
    cuz ksp
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    (Original post by bwr19)
    How is the answer to this C and not B? Surely

    T = 2\pi\sqrt{\frac{l}{2g}}

    so

    \sqrt{2}T = 2\pi\sqrt{\frac{l}{g}}


    ????Name:  Capture.PNG
Views: 129
Size:  37.6 KB
    ah dude thats not how it works. This isnt like maths where you resolve the equation. Its more like if the right hand side of the equation changes by a factor of

    1/(root2) then the other side will also go down by 1/(root2)
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    (Original post by Kennethm)
    ah dude thats not how it works. This isnt like maths where you resolve the equation. Its more like if the right hand side of the equation changes by a factor of

    1/(root2) then the other side will also go down by 1/(root2)
    Good spot.

    I made the same mistake...
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    (Original post by Kennethm)
    ah dude thats not how it works. This isnt like maths where you resolve the equation. Its more like if the right hand side of the equation changes by a factor of

    1/(root2) then the other side will also go down by 1/(root2)
    Haha I've realised this now after making the same mistake on a similar question. Thanks for the help.
 
 
 
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